Question

lf $A_1\left(z_1\right)$ , $A_2\left({\overline{z}}_1\right)$ are the adjacent vertices of a regular polygon. lf $\frac{Im\left(\overline{z_1}\right)}{Re\left(z_1\right)}=1-\sqrt{2}$

then number of sides of the polygon is equal to?

• A. $6$
• B. $8$
• C. $16$
• D. $12$

Answer 1

Answer: (B)

$8$

Let $z_1=a+ib$
$\overline{z}=a-ib$
$\therefore \frac{Im{z}_1}{Re{z}_1}=\frac{-b}{a}=1-\sqrt{2}$
$\therefore \frac{b}{a}=-(1-\sqrt{2})=\sqrt{2}$
$\therefore arg{z}_1=ta{n}^{-1}\left(\frac{b}{a}\right)=ta{n}^{-1}(\sqrt{2}-1)=\frac{\pi }{8}\left(\frac{{45}^0}{2}\right)$
$\therefore$ As $z_1$ & $\overline{z_1}$ are adjacent vertices, so we have
$\therefore$ each side of the polygon subtends an angle of ${}^{\prime }2{\theta }^{\prime }$ at
the centre of polygon.
$\therefore 2\theta =2arg{z}_1=2\times \frac{n}{8}=\frac{\pi }{4}.$
$\therefore$ Total sides of polygon $=\frac{2\pi }{\frac{\pi }{4}}=2\times 4=8$